$\rightarrow I_a =$The $A-$excenter of $\triangle{ABC}$. $\rightarrow \ell =$The line tangent to $k(I,r)$ at $L$. $\rightarrow (AC,k(I,r))=E$ , $(AB,k(I,r))=F$. $\rightarrow (BC,EF,\ell)=H$. $\rightarrow (I_aM,LH)=K$. $\rightarrow IQ \perp AD$ , $Q \in AD$. $\rightarrow I_aD' \perp BC$ , $D' \in BC$. $\rightarrow$ The facts that $BC,EF,\ell$ are concurrent and $I_aM \parallel AD$ and $HB.HC=HD.HM$are well-known, so I don't post a proof for them unless needed. $\rightarrow HD=HL$ $,$ $DL \parallel MK \Rightarrow HL.HK=HD.HM=HB.HC \Rightarrow BLKC$ is cyclic. Now the only thing we have to prove is that $KM.MN=BM.CM$. $\rightarrow MI_a \parallel AD \Rightarrow \triangle{IQD} \sim \triangle{MD'I_a} \Rightarrow \frac{I_aM}{r}=\frac{2r_a}{DL}=\frac{2r_a.HM}{MK.HD}$ $\Rightarrow \frac{MK.I_aM}{2}= \frac{r.r_a.HM}{HD} = \frac{BD.CD.HM}{HD} = \frac{BD.BM.HC}{HD} = BM.CM \blacksquare$(Note that in the two last conclusions, we used the fact that $HB.HC=HD.HM$.)

My solution: Let $ D', D'' $ be the reflection of $ D, D' $ with respect to $ M, K $. Let $ X $ be the intersection of the polar of $ A $ with respect to $ (I) $ and $ BC $ . Let $ N' $ be the reflection of $ N $ with respect to the perpendicular bisector of $ BC $ . By homothety with center $ A $ we get $ A, L, D, D'' $ are collinear. Since it's well known that $ KD'\perp BC $ , so we get $ N' $ lie on $ AD'' $ . Since $ AD $ is the polar of $ X $ with respect to $ (I) $ , so we get $ XL $ is tangent to $ (I) $ and $ (X,D;B,C)=-1 $ . Since $ XL=XD $ and $ N'M=N'D $ , so $ \angle XLD =\angle LDX =\angle N'DM =\angle DMN' $ . i.e. $ L,X,M,N' $ are concyclic Since $ DL \cdot DN'=DX \cdot DM=DB \cdot DC $ , so we get $ B, C, N, N', L $ are concyclic. Q.E.D

Let $E$, $F$ be points of tangency of the incircle with $AC$ and $AB$. It is easy to show that $MK \parallel AD$. Besides $FD \parallel BK$. Hence we get $\angle KBM= \angle BDF = \angle FLD$ and $ \angle FDL = \angle BKM$, thus $\Delta FDL \sim \Delta BKM$. Notice that $BL$ is a symmedian of the triangle $FDL$ so $\angle BLF= \angle GLD= \angle KBN$, where $G$ is a midpoint of $FD$. From that we get that $\angle NBM= \angle BLD$. Analogously $\angle DLC=\angle NCM$, thus $B, N, C, L$ are concyclic.

Pinionrzek wrote: It is easy to show that $MK \parallel AD$. Can someone show this?

PatrikP wrote: Pinionrzek wrote: It is easy to show that $MK \parallel AD$. Can someone show this? Note that the homothety mapping the incircle to the excircle maps $D$ to the antipode of $E$.This with $DM=EM$ makes it clear.

Let $\ell$ be the perpendicular bisector of $BC$. Define points $K'$ and $N'$ as the images of $K$ and $N$, respectively in the symmetry about $\ell$. Let $T$ be a point on line $BC$ such that $(B, C; D, T)=-1$ and note that $TL$ is tangent to $\omega$. It is well-known that $B, I, C, K$ are concyclic, so $B, I, C, K'$ are also concyclic. By power of point we get $$DI\cdot DK'=DB\cdot DC=DM\cdot DT,$$so $T, I, M, K'$ are concyclic. Thus, $$\angle (AD, BC)=\angle TDL=\angle TID=\angle TMK'=\angle N'DM,$$so $A, D, N'$ are collinear. Finally, observe that $\angle TLN'=\angle TMN'$, so $T, L, M, N'$ are concyclic. By power of point, $$DB\cdot DC=DT\cdot DM=DL\cdot DN',$$so $B, L, C, N'$ are concyclic. The result follows as $BCNN'$ is an isosceles trapezoid. $\square$

$$AD||MK$$triangle $FQD$ similar triangle $BMK$. $$ \angle BQD=\angle NBM $$Similarity:$$\angle CQD=\angle NCM$$$B,Q,C,N$ concyclic.

Same solution as #1 but bit more motivated Claim 1: $MK \parallel AD$ Let $A$-excircle touch $BC$ at $D'$ and $R$ be it's antipode in excircle. Then by homothety at $A$ sending incircle to excircle sends $D$ to $R$ hence $\overline{A-D-R}$. $M$ is midpoint of $DD'$ thus by MPT in $\triangle DD'R$ we get $MK \parallel AD$. Let $S$ be reflection of $M$ above $TI$ Claim 2: $S \in (LBC)$ Let $E,F$ be other touch point of incircle with $AC,AB$ then $T =EF \cap BC$. $A$ lie on polar of $T$, therefore $TL$ tangent to incircle. Note we have $MK \perp TI$. Perpendicular bisector of $MS$ and $DL$ is $TI$ only. This give us $L,S,M,D$ cyclic. As $(T,D;B,C)=-1$ by using power of point we get $$TB \cdot TC =TM \cdot TD = TL \cdot TS$$Which give us $S \in (LBC)$. Now the motivation for introducing $S$ come in picture. We claim $N$ lie on $(SLBC)$. Note $\angle IXK = 90$ hence $X \in (IBKC)$. We use power of point at $M$ $$MB\cdot MC = MX \cdot MK = 2\cdot MX \cdot \frac{MK}{2}= MS \cdot MN$$which yields $N \in (SBC)$. Hence $L,B,C,N$ cyclic.